Integrand size = 27, antiderivative size = 89 \[ \int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^4 d}-\frac {\left (a^2-b^2\right ) \sin (c+d x)}{b^3 d}+\frac {a \sin ^2(c+d x)}{2 b^2 d}-\frac {\sin ^3(c+d x)}{3 b d} \] Output:
a*(a^2-b^2)*ln(a+b*sin(d*x+c))/b^4/d-(a^2-b^2)*sin(d*x+c)/b^3/d+1/2*a*sin( d*x+c)^2/b^2/d-1/3*sin(d*x+c)^3/b/d
Time = 0.15 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {6 a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))+6 b \left (-a^2+b^2\right ) \sin (c+d x)+3 a b^2 \sin ^2(c+d x)-2 b^3 \sin ^3(c+d x)}{6 b^4 d} \] Input:
Integrate[(Cos[c + d*x]^3*Sin[c + d*x])/(a + b*Sin[c + d*x]),x]
Output:
(6*a*(a^2 - b^2)*Log[a + b*Sin[c + d*x]] + 6*b*(-a^2 + b^2)*Sin[c + d*x] + 3*a*b^2*Sin[c + d*x]^2 - 2*b^3*Sin[c + d*x]^3)/(6*b^4*d)
Time = 0.30 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3316, 27, 522, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sin (c+d x) \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x) \cos (c+d x)^3}{a+b \sin (c+d x)}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {\int \frac {\sin (c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )}{a+b \sin (c+d x)}d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {b \sin (c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )}{a+b \sin (c+d x)}d(b \sin (c+d x))}{b^4 d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {\int \left (-\left (\left (1-\frac {b^2}{a^2}\right ) a^2\right )+b \sin (c+d x) a-b^2 \sin ^2(c+d x)+\frac {a^3-a b^2}{a+b \sin (c+d x)}\right )d(b \sin (c+d x))}{b^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-b \left (a^2-b^2\right ) \sin (c+d x)+a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))+\frac {1}{2} a b^2 \sin ^2(c+d x)-\frac {1}{3} b^3 \sin ^3(c+d x)}{b^4 d}\) |
Input:
Int[(Cos[c + d*x]^3*Sin[c + d*x])/(a + b*Sin[c + d*x]),x]
Output:
(a*(a^2 - b^2)*Log[a + b*Sin[c + d*x]] - b*(a^2 - b^2)*Sin[c + d*x] + (a*b ^2*Sin[c + d*x]^2)/2 - (b^3*Sin[c + d*x]^3)/3)/(b^4*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 0.52 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {-\frac {\frac {\sin \left (d x +c \right )^{3} b^{2}}{3}-\frac {b a \sin \left (d x +c \right )^{2}}{2}+a^{2} \sin \left (d x +c \right )-\sin \left (d x +c \right ) b^{2}}{b^{3}}+\frac {a \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4}}}{d}\) | \(83\) |
default | \(\frac {-\frac {\frac {\sin \left (d x +c \right )^{3} b^{2}}{3}-\frac {b a \sin \left (d x +c \right )^{2}}{2}+a^{2} \sin \left (d x +c \right )-\sin \left (d x +c \right ) b^{2}}{b^{3}}+\frac {a \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4}}}{d}\) | \(83\) |
parallelrisch | \(-\frac {12 a^{3} \left (-\ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )\right )-12 a \,b^{2} \left (-\ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )\right )-3 a \,b^{2}+12 \sin \left (d x +c \right ) a^{2} b +3 a \,b^{2} \cos \left (2 d x +2 c \right )-9 \sin \left (d x +c \right ) b^{3}-b^{3} \sin \left (3 d x +3 c \right )}{12 d \,b^{4}}\) | \(163\) |
risch | \(-\frac {i a^{3} x}{b^{4}}+\frac {i x a}{b^{2}}-\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{2} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 b^{3} d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )}}{8 b d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 b^{3} d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )}}{8 b d}-\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 b^{2} d}-\frac {2 i a^{3} c}{b^{4} d}+\frac {2 i a c}{b^{2} d}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{4} d}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{2} d}+\frac {\sin \left (3 d x +3 c \right )}{12 b d}\) | \(250\) |
norman | \(\frac {-\frac {2 \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{3} d}-\frac {2 \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{b^{3} d}-\frac {2 \left (9 a^{2}-5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 b^{3} d}-\frac {2 \left (9 a^{2}-5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 b^{3} d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b^{2} d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{b^{2} d}+\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d \,b^{2}}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {a \left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{4} d}-\frac {a \left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{b^{4} d}\) | \(270\) |
Input:
int(cos(d*x+c)^3*sin(d*x+c)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/b^3*(1/3*sin(d*x+c)^3*b^2-1/2*b*a*sin(d*x+c)^2+a^2*sin(d*x+c)-sin( d*x+c)*b^2)+a*(a^2-b^2)/b^4*ln(a+b*sin(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {3 \, a b^{2} \cos \left (d x + c\right )^{2} - 6 \, {\left (a^{3} - a b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \, {\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (d x + c\right )}{6 \, b^{4} d} \] Input:
integrate(cos(d*x+c)^3*sin(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/6*(3*a*b^2*cos(d*x + c)^2 - 6*(a^3 - a*b^2)*log(b*sin(d*x + c) + a) - 2 *(b^3*cos(d*x + c)^2 - 3*a^2*b + 2*b^3)*sin(d*x + c))/(b^4*d)
Timed out. \[ \int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3*sin(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, b^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{2} + 6 \, {\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}{b^{3}} - \frac {6 \, {\left (a^{3} - a b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{4}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*sin(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/6*((2*b^2*sin(d*x + c)^3 - 3*a*b*sin(d*x + c)^2 + 6*(a^2 - b^2)*sin(d*x + c))/b^3 - 6*(a^3 - a*b^2)*log(b*sin(d*x + c) + a)/b^4)/d
Time = 0.17 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {{\left (a^{3} - a b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{4} d} - \frac {2 \, b^{2} d^{2} \sin \left (d x + c\right )^{3} - 3 \, a b d^{2} \sin \left (d x + c\right )^{2} + 6 \, a^{2} d^{2} \sin \left (d x + c\right ) - 6 \, b^{2} d^{2} \sin \left (d x + c\right )}{6 \, b^{3} d^{3}} \] Input:
integrate(cos(d*x+c)^3*sin(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
(a^3 - a*b^2)*log(abs(b*sin(d*x + c) + a))/(b^4*d) - 1/6*(2*b^2*d^2*sin(d* x + c)^3 - 3*a*b*d^2*sin(d*x + c)^2 + 6*a^2*d^2*sin(d*x + c) - 6*b^2*d^2*s in(d*x + c))/(b^3*d^3)
Time = 19.66 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\sin \left (c+d\,x\right )\,\left (\frac {1}{b}-\frac {a^2}{b^3}\right )-\frac {{\sin \left (c+d\,x\right )}^3}{3\,b}+\frac {a\,{\sin \left (c+d\,x\right )}^2}{2\,b^2}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (a\,b^2-a^3\right )}{b^4}}{d} \] Input:
int((cos(c + d*x)^3*sin(c + d*x))/(a + b*sin(c + d*x)),x)
Output:
(sin(c + d*x)*(1/b - a^2/b^3) - sin(c + d*x)^3/(3*b) + (a*sin(c + d*x)^2)/ (2*b^2) - (log(a + b*sin(c + d*x))*(a*b^2 - a^3))/b^4)/d
Time = 0.18 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.92 \[ \int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{3}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a \,b^{2}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a^{3}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a \,b^{2}-2 \sin \left (d x +c \right )^{3} b^{3}+3 \sin \left (d x +c \right )^{2} a \,b^{2}-6 \sin \left (d x +c \right ) a^{2} b +6 \sin \left (d x +c \right ) b^{3}-4 a \,b^{2}}{6 b^{4} d} \] Input:
int(cos(d*x+c)^3*sin(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
( - 6*log(tan((c + d*x)/2)**2 + 1)*a**3 + 6*log(tan((c + d*x)/2)**2 + 1)*a *b**2 + 6*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**3 - 6*l og(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a*b**2 - 2*sin(c + d* x)**3*b**3 + 3*sin(c + d*x)**2*a*b**2 - 6*sin(c + d*x)*a**2*b + 6*sin(c + d*x)*b**3 - 4*a*b**2)/(6*b**4*d)